Two loudspeakers, A and B are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. The frequency of the sound waves produced by the loudspeakers is 206 Hz. Consider a point P between the speakers and along the line connecting them, a distance x to the right of A. Both speakers emit sound waves that travel directly from the speaker to point P. For what values of x will: a. destructive interference occur at P
b. constructive interference occur at P

a) 0, ±1.65  b)  ± 0.825m

Explanation:

This is a sound interference exercise, it can be described by the path difference between the two waves

for the case of constructive interference

          Δr = 2n λ/ 2                   n = 0, 1, 2 ...

for the case of destructive interference

          Δr = (2n + 1) λ/ 2

the speed of sound is related to the wavelength

          v = λ f

          λ = v / f

          λ= 340/206

          λ = 1.65 m

let's set a reference system in the center between the two speakers

a) let's find the distances for constructive interference

            Δr = 2n 1.65 / 2

            Δr = 1.65 n

* the first interference occurs at n = 0

          Δr = 0

therefore the interference in the center is maximum

* n = 1

          Dr = 1.65 m

the second inference occurs at 1.65 m from the center, therefore there is a right wing and a left wing,

We do not have any more interference between the speakers because

n = 2 Δr = 3.3m this distance can be from the speaker

b) let's look for the destructive interference points

           Δr = (2n + 1) 1.65 / 2

           Δr = (2n + 1) 0.825

m = 0       Δr = 0.825m

m = 1        Δr = 2,475m

We can see that we only have the first destructive interference, one on each side.

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the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s